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I just found the Laurent series for $z^{2}e^{1/z}$ for $z = 0$, and now I need to find it at $z = \infty$. (for $z=0$, it was $\displaystyle \sum_{n=0}^{\infty}\frac{z^{2-n}}{n!}$, by the way).

I'm not sure how to approach this problem for $z = \infty$, however...

I came across one method involving finding the series for $\zeta^{-2}(f(\zeta^{-1}))$. Apparently, this gives the series at $\infty$, but I don't really understand how.

Please help - full, detailed solution preferred. I am extremely confused and am trying to teach myself this stuff; it helps me to see fully worked out examples. Thank you.

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  • $\begingroup$ @robjohn, I feel really awful asking you this after all the help you've given me today. But, you're the man when it comes to complex! $\endgroup$
    – user100463
    Apr 20, 2016 at 23:25
  • $\begingroup$ generally, the Laurent series of $f(z)$ at $\infty$ simply means, with $g(z) = f(1/z)$ and $\sum_k c_k z^k$ the Laurent series of $g(z)$ at $z=0$, that the answer is $\sum_k c_k (1/z)^{k}$... $\endgroup$
    – reuns
    Apr 21, 2016 at 0:22

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Laurent series at $z=\infty$ is just a Laurent series in $\frac1z$

Just use the standard series for $e^x$ and set $x=1/z$: $$ \begin{align} z^2e^{1/z} &=z^2\overbrace{\sum_{k=0}^\infty\frac1{z^kk!}}^{\text{standard series}}\\ &=\sum_{k=0}^\infty\frac1{z^{k-2}k!} \end{align} $$ If you want to be pedantic and write it as a series in $\frac1z$, you can write this as $$ \frac1{\left(\frac1z\right)^2}+\frac1{\frac1z}+\sum_{k=0}^\infty\frac{\left(\frac1z\right)^k}{(k+2)!} $$

Mathematica concurs

Simply as a verification, I ran this through Mathematica and it gave

Series[z^2Exp[1/z],{z,Infinity,5}] $$ z^2+z+\frac12+\frac1{6z}+\frac1{24z^2}+\frac1{120z^3}+\frac1{720z^4}+\frac1{5040z^5}+O\left(\frac1z\right)^6 $$

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  • $\begingroup$ This is the Laurent series at $z=0$. $\endgroup$
    – Rick Sanchez
    Apr 20, 2016 at 23:34
  • $\begingroup$ @robjohn, yeah I think you got it backwards. This was the problem I was having, too. I was wondering if you tweak it though to make it what I'm looking for, though? $\endgroup$
    – user100463
    Apr 20, 2016 at 23:34
  • $\begingroup$ This is the Laurent series at infinity. The two are the same when the series converges everywhere (except at $0$ and $\infty$). You just have to write it as a series in $\frac1z$. $\endgroup$
    – robjohn ♦
    Apr 20, 2016 at 23:43
  • $\begingroup$ think of it this way: what is the Laurent expansion of $\frac{e^z}{z^2}$ at $z=0$? Substitute $z\mapsto\frac1z$, $\endgroup$
    – robjohn ♦
    Apr 21, 2016 at 0:37

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